Problem: Find $\dfrac{d}{dx}\left(\dfrac{\ln(x)}{e^x}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1-x\ln(x)}{xe^x}$ (Choice B) B $\dfrac{1}{xe^x}$ (Choice C) C $\dfrac{1-xe^x}{xe^{2x}}$ (Choice D) D $\dfrac{1}{x}-e^x$
Answer: $\dfrac{\ln(x)}{e^x}$ is the quotient of two, more basic, expressions: $\ln(x)$ and $e^x$. Therefore, the derivative of the expression can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d d x ( ln ( x ) e x ) = d d x ( ln ( x ) ) e x − ln ( x ) d d x ( e x ) ( e x ) 2 = 1 x ⋅ e x − ln ( x ) ⋅ e x ( e x ) 2 = e x ( 1 x − ln ( x ) ) ( e x ) 2 = 1 − x ln ( x ) x e x The quotient rule Differentiate ln ( x ) and e x Simplify Simplify \begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{\ln(x)}{e^x}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\ln(x))e^x-\ln(x)\dfrac{d}{dx}(e^x)}{(e^x)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{\dfrac1x\cdot e^x-\ln(x)\cdot e^x}{(e^x)^2}&&\gray{\text{Differentiate }\ln(x)\text{ and }e^x} \\\\ &=\dfrac{\cancel{e^x}\left(\dfrac1x-\ln(x)\right)}{(e^x)^\cancel2}&&\gray{\text{Simplify}} \\\\ &=\dfrac{1-x\ln(x)}{xe^x}&&\gray{\text{Simplify}} \end{aligned} In conclusion, $\dfrac{d}{dx}\left(\dfrac{\ln(x)}{e^x}\right)=\dfrac{1-x\ln(x)}{xe^x}$